Min + Max? (Trigonometry)

$\begin{aligned} \sin^2 x + \sin^2 y & = \frac 12 \\ \frac {1-\cos 2x}2 + \frac {1-\cos 2y}2 & = \frac 12 \\ 1-\cos 2x + 1-\cos 2y & = 1 \\ \cos 2x + \cos 2y & = 1 & \small \color{#3D99F6} \text{Using Weierstrass substitution} \\ \frac {1-\tan^2 x}{1+\tan^2 x} + \frac {1-\tan^2 y}{1+\tan^2 y} & = 1 \\ (1-\tan^2 x)(1+\tan^2 y) + (1-\tan^2 y)(1+\tan^2 x) & = (1+\tan^2 x)(1+\tan^2 y) \\ \implies S = \tan^2 x + \tan^2 y & = 1 - 3{\color{#3D99F6}\tan^2 x \tan^2 y} \end{aligned}$

Note that $\sin^2 x + \sin^2 y = \frac 12$ , $\implies 0 \le \sin^2 x, \sin^2 y \le \frac 12$ . And that $S$ is minimum or $S=m$ when ${\color{#3D99F6}\tan^2 x \tan^2 y}$ is maximum and $S$ is maximum or $S=M$ when ${\color{#3D99F6}\tan^2 x \tan^2 y}$ is minimum.

Using AM-GM inequality :

$\begin{aligned} \tan x + \tan y & \ge 2 \sqrt{\tan x \tan y} \\ \implies \tan^2 x \tan^2 y & \le \frac {(\tan x + \tan y)^4}{16} \small \color{#3D99F6} \text{Equality occurs when }x=y = \frac \pi 6 \text{ (see note).} \\ & \le \frac {\left(\frac 2{\sqrt 3} \right)^4}{16} = \color{#3D99F6} \frac 19 \\ \implies m & = 1 - 3 \times {\color{#3D99F6} \frac 19} = \frac 23 \end{aligned}$

Note: Equality occurs when $\tan x = \tan y$ or $x=y$ and $\sin^2 x + \sin^2 x = \frac 12$ , $\implies \sin x = \frac 12$ or $x = \frac \pi 6$ .

Minimum ${\color{#3D99F6}\tan^2 x \tan^2 y}$ is 0, when either $x = 0$ or $y=0$ . $\implies M = 1 - 3({\color{#3D99F6}0}) = 1$ .

Therefore, $m + M = \dfrac 23 + 1 = \boxed{\dfrac 53}$

---

Previous solution

$\begin{aligned} \sin^2 x + \sin^2 y & = \frac 12 \\ \implies \sin^2 y & = \frac 12 - \sin^2 x \\ \implies \cos^2 y & = 1 - \frac 12 + \sin^2 x = \frac 12 + \sin^2 x \end{aligned}$

Then, we have:

$\begin{aligned} S & = \tan^2 x + \tan^2 y \\ & = \frac {\sin^2 x}{\cos^2 x} + \frac {\sin^2 y}{\cos^2 y} & \small \color{#3D99F6} \text{Let }u=\sin^2 x \\ & = \frac u{1-u} + \frac {\frac 12 - u}{\frac 12+u} \\ & = \frac u{1-u} + \frac {1 - 2u}{1+2u} \\ & = \frac {1-2u+4u^2}{1+u-2u^2} \\ & = \frac {3-2(1 +u-2u^2)}{1+u-2u^2} \\ & = \frac 3{1+u-2u^2}-2 \\ & = \frac 3{\color{#3D99F6}\frac 98 - 2\left(u-\frac 14\right)^2}-2 \end{aligned}$

We note that $S$ is minimum or $S=m$ when $\color{#3D99F6}\frac 98 - 2\left(u-\frac 14\right)^2$ is maximum and $S$ is maximum or $S=M$ when $\color{#3D99F6}\frac 98 - 2\left(u-\frac 14\right)^2$ is minimum. Note that $\sin^2 x + \sin^2 y = \frac 12$ , $\implies 0 \le \sin^2 x, \sin^2 y \le \frac 12$ , $\implies 0 \le u \le \frac 12$ . Therefore,

$\begin{aligned} m & = \frac 3{\frac 98 - 2\left({\color{#3D99F6}0}\right)^2}-2 & \small \color{#3D99F6} \text{Note that } \frac 98 - 2\left(u-\frac 14\right)^2 \text{ is maximum when }u = \frac 14 \\ & = \frac {3\times 8}9 - 2 \\ & = \frac 23 \end{aligned}$

$\begin{aligned} M & = \frac 3{\frac 98 - 2\left({\color{#3D99F6}\frac 12-\frac 14}\right)^2}-2 & \small \color{#3D99F6} \text{Note that } \frac 98 - 2\left(u-\frac 14\right)^2 \text{ is maximum when }u = \frac 12 \\ & = \frac 3{\frac 98 - \frac 18} - 2 \\ & = 1 \end{aligned}$

$\implies m+M = \dfrac 23 + 1 = \boxed{\dfrac 53}$

We have to find the range of $S$ such that the system of equations below has some solution

$\begin{cases} \sin ^{ 2 }{ x } +\sin ^{ 2 }{ y } =\frac { 1 }{ 2 } \\ \tan ^{ 2 }{ x } +\tan ^{ 2 }{ y } =S \end{cases}\quad \quad \quad \quad \left( 1 \right)$

From $(1)$ we have:

$\begin{cases} \cos ^{ 2 }{ x } +\cos ^{ 2 }{ y } =\frac { 3 }{ 2 } \\ \frac { 1 }{ \cos ^{ 2 }{ x } } +\frac { 1 }{ \cos ^{ 2 }{ y } } =S+2 \end{cases}\quad \left( \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } =\sin ^{ 2 }{ y } +\cos ^{ 2 }{ y } =1 \right)$

$\Longrightarrow \quad \frac { \left( \cos ^{ 2 }{ x } +\cos ^{ 2 }{ y } \right) }{ \cos ^{ 2 }{ x } \cos ^{ 2 }{ y } } =S+2\\ \Longrightarrow \quad \cos ^{ 2 }{ x } \cos ^{ 2 }{ y } =\frac { 3 }{ 2\left( S+2 \right) }$

So, $\cos ^{ 2 }{ x }, \cos ^{ 2 }{ y }$ are the roots of the equation: $f\left( X \right) ={ X }^{ 2 }-\frac { 3 }{ 2 } X+\frac { 3 }{ 2\left( S+2 \right) } =0$

Since $0\le \cos ^{ 2 }{ x } \le 1 , 0\le \cos ^{ 2 }{ y } \le 1$ , all the zeros of $f\left( X \right)$ satisfy $0\le X\le 1$

To achieve that,

$\begin{cases} \Delta =9-\frac { 24 }{ S+2 } \ge 0 \\ af(0)=\frac { 3 }{ S+2 } >0 \\ \frac { -b }{ 2a } -0>0 \\ af(1)=\frac { 3 }{ S+2 } -1\ge 0 \\ \frac { -b }{ 2a } -1<0 \end{cases}$

$\Longleftrightarrow \begin{cases} \frac { 9S-6 }{ S+2 } \ge 0 \\ S\ge -2 \\ \frac { -S+1 }{ S+2 } \ge 0 \end{cases}\\ \Longleftrightarrow \begin{cases} S\ge 2/3\quad or\quad S\le -2 \\ S>-2 \\ -2\le S\le 1 \end{cases}\\ \Longleftrightarrow 2/3\le S\le 1.$

Therefore, $m=\frac { 2 }{ 3 } ,\quad M=1,\quad m+M=\boxed { \frac { 5 }{ 3 } } .$